Intuition The one core idea
An alkane is a chain of carbon atoms wearing hydrogen atoms, held by strong plain single bonds — so almost nothing attacks it except a lone-electron fragment called a free radical . Everything in this chapter is either a way to join carbon pieces into a chain (preparation) or a way to let a radical swap a hydrogen for a halogen (halogenation).
Before you touch Wurtz, Kolbe, or halogenation, you must be able to read every squiggle the parent note throws at you. Below, each symbol is built from nothing: plain words first, then the picture, then why the topic can't live without it .
Definition The chemical letters
C , H , X , N a
C = one carbon atom. Picture a ball that wants to hold four hands (bonds).
H = one hydrogen atom. A tiny ball that holds only one hand.
X = a stand-in letter for any halogen — F (fluorine), C l (chlorine), B r (bromine), I (iodine). We write X when the argument works for all of them.
N a = a sodium metal atom, eager to give away one electron.
The dash in C – C or C – H is a single bond : two electrons shared between two atoms, one from each. Picture two balls holding one shared pair of hands.
Intuition Why "four hands" for carbon matters
Every carbon in an alkane uses all four hands for single bonds only — no leftover pairs, no double bonds. That "all hands full, all single" state has a name: saturated . A saturated carbon has nothing sticking out to grab a passing reagent, which is exactly why alkanes are unreactive.
Definition Subscripts and the general formula
In C 2 H 6 , the small numbers (subscripts ) count atoms: 2 carbons, 6 hydrogens .
The general alkane formula is
C n H 2 n + 2
Here n is just "how many carbons " — pick any whole number (n = 1 , 2 , 3 , … ). Then the hydrogen count is forced to be 2 n + 2 .
WHY that exact count 2 n + 2 , and why do we need a formula at all? Because it is the fingerprint of "fully saturated." Line up n carbons in a chain: the two end carbons each have 3 spare hands, and each middle carbon has 2 spare hands. Count them and you always land on 2 n + 2 hydrogens. If a molecule has fewer H than 2 n + 2 , some carbons must be sharing a double bond — it is not an alkane. So the formula is your test for "is this thing saturated?"
Worked example Check the formula
Propane has n = 3 carbons, so 2 n + 2 = 2 ( 3 ) + 2 = 8 hydrogens → C 3 H 8 . Butane: n = 4 ⇒ 2 ( 4 ) + 2 = 10 ⇒ C 4 H 10 .
R
R means "any alkyl group " — a carbon chain with one hand free, waiting to attach. Think of R as "whatever is left after we chop off the reacting bit ."
Picture C H 3 – C H 2 – C l . The interesting part is the – C l ; everything to its left (C H 3 C H 2 ) we lazily call R . So the whole thing is R – C l , or in symbol soup R – X .
Intuition Why chemists invented
R
Wurtz, Kolbe, and halogenation all work the same way no matter which chain you plug in. Writing R lets us state one equation that covers ethyl, propyl, butyl… at once. When you see 2 R – X + 2 N a → R – R , mentally replace R with a real group like C H 3 C H 2 to make it concrete.
s p 3 and the tetrahedron
s p 3 is a label for how carbon aims its four hands : not flat, but pointing to the four corners of a tetrahedron (a triangular pyramid), roughly 109. 5 ∘ apart.
WHY the topic mentions s p 3 at all: it explains two later facts. (1) The chain is a zig-zag , not a straight line, which is why a branched alkane can curl into a compact near-sphere (matters for boiling points in §Properties of the parent). (2) There are no flat π -electron clouds sticking above the atoms for an attacker to grab — reinforcing why alkanes are inert. You don't need to derive s p 3 here; just picture the pyramid. Deeper electron-donation effects live in Hyperconjugation .
∙ = one unpaired electron
A superscript dot, as in C l ∙ or C H 3 ∙ , marks an atom or group with one lonely, unpaired electron — a bond that got split down the middle so each side kept one of the two shared electrons. Such a species is a free radical .
Compare the three ways a bond can break, side by side:
Homolysis (even split): A : B → A ∙ + B ∙ . Each keeps one electron. This makes radicals .
Heterolysis (uneven split): A : B → A + + : B − . One side keeps both electrons. This makes ions — a positive A + and a negative B − (with a lone pair, drawn : ).
Intuition Why radicals are the star of this chapter
Ionic reagents attack polar or electron-rich spots. Alkanes have none. A radical is electron-hungry and neutral , so it can rip an H straight off a plain C–H bond — the only thing that can. Every preparation mechanism (Wurtz, Kolbe) and all of halogenation runs through radicals. Master the dot and you've unlocked the chapter. Details of radical shape and lifetime: Free radicals and reaction intermediates .
+ , − and the reaction arrow →
N a + = sodium that lost one electron (now positive). X − = halogen that gained one electron (now negative). These charged particles are ions .
A plain arrow → means "turns into ." The left side (reactants) becomes the right side (products).
Writing above the arrow, like dry ether or h ν or N i , Δ , names the conditions — solvent, light, catalyst, or heat.
Definition The special condition symbols
h ν = a packet of light energy (a photon). We shine it to snap a weak bond apart. (h and ν are physics constants for photon energy — here just read "light.")
Δ (Greek capital delta) = "apply heat ."
e − = a single free electron , the thing that flows in electricity.
WHY these live above the arrow: the same reactants can do nothing or explode depending on conditions. C H 4 + C l 2 sit quietly in the dark; add h ν and they react. The condition is half the reaction.
Definition Words you need for Kolbe
Electrolysis = pushing electricity through a liquid to force a reaction. See Electrolysis and redox at electrodes .
Anode = the positive electrode, where particles lose electrons (this loss is called oxidation ).
R C O O − = a carboxylate ion — the leftover after a carboxylic acid R C O O H gives up its acidic H + . Picture R – C ( = O ) – O − .
C O 2 = carbon dioxide, an extremely stable gas. Its stability is the prize that drives Kolbe: losing C O 2 is downhill.
WHY the topic needs "anode = oxidation = lose electron": Kolbe only works because the carboxylate is dragged to the anode and stripped of an electron, becoming a radical R C O O ∙ . Without the electrode vocabulary, step 2 of the mechanism is unreadable.
Intuition Symmetrical vs mixed
Symmetrical alkane = same group on both sides of the new bond, e.g. C H 3 C H 2 – C H 2 C H 3 (both halves are ethyl). Wurtz and Kolbe only cleanly make symmetrical products, because two identical radicals join. Mix two different pieces and they cross-couple into a mixture — three products, hard to separate. Hold this picture now; the parent uses it twice.
Intuition Selectivity vs reactivity (preview)
Reactive = grabs almost any H, fast and indiscriminate (chlorine). Selective = fussy, prefers the H that yields the most stable radical (bromine). These are opposite qualities — the parent's stability order 3 ∘ > 2 ∘ > 1 ∘ decides which H a selective reagent picks. Why some radicals are steadier: Hyperconjugation .
Atoms C H X Na and the single bond
Formula CnH2n+2 saturated
sp3 tetrahedron and chain shape
Bond breaking homolysis vs heterolysis
Radical dot one unpaired electron
Ions plus and minus charges
Free-radical halogenation
Kolbe anode oxidation CO2
Wurtz couple two R groups
Physical properties boiling point
Selectivity 3 over 2 over 1
I can read C n H 2 n + 2 and find the H count for any n Yes — H = 2 n + 2 ; e.g. n = 5 gives C 5 H 12 .
I know what the dash in C – C physically means A single bond = one shared pair of electrons.
I can say what R and X stand for R = any alkyl group (rest of the chain); X = any halogen F , C l , B r , I .
I can explain the bullet ∙ A free radical — one unpaired electron from an even (homolytic) bond split.
I can tell homolysis from heterolysis Homolysis splits evenly → two radicals; heterolysis splits unevenly → two ions.
I know why alkanes are unreactive Saturated, non-polar, no π -electrons or lone pairs — only radicals can attack.
I can read h ν , Δ , e − Light (photon), heat, and a single electron respectively.
I know what happens at the anode Oxidation — particles lose electrons (needed for Kolbe).
I know why C O 2 leaving drives Kolbe C O 2 is very stable; forming it is energetically downhill.
I can define symmetrical vs mixed products Symmetrical = same group both sides (clean); mixed = cross-coupling gives a messy mixture.